Solveeit Logo

Question

Question: 200ml of \({{O}_{2}}\) gas maintained at 700mm pressure and 250ml of \({{N}_{2}}\) gas maintained at...

200ml of O2{{O}_{2}} gas maintained at 700mm pressure and 250ml of N2{{N}_{2}} gas maintained at 720mm pressure are put together in a one litre flask. If the temperature is kept constant, the final pressure of the mixture is:
[A] 450
[B] 320
[C] 632
[D] 316

Explanation

Solution

To solve this, you need to consider the ideal gas equation. Remember that the temperature is constant and calculate the number of moles of each of the given gases. As we have mixed the gases in a closed container, the number of moles of gas in the mixture will be equal to the sum of the number of moles of each gas. From here, you can calculate the final pressure of the mixture.

Complete step by step solution:
The ideal gas law equation is an equation of state variables of a hypothetical ideal gas. We know that the ideal gas equation is – PV=nRTPV = nRT where, P is the pressure, V is the volume of the gas, n is the number of moles, R is the universal gas constant which has a fixed value and T is the temperature.
Now, in the question we have oxygen and nitrogen gases.
So, firstly let us consider oxygen gas.
Pressure = 700 mm
Volume = 200 ml = 0.2 L
T is constant.
We know, PV=nRTPV = nRT
Since T is constant, so n=n = PVR\dfrac{PV}{R}
So, putting the values, we will have number of moles of oxygen gas initially= 700×0.2R\dfrac{700\times 0.2}{R}
Similarly, for nitrogen gas, P = 720 mm and V = 250 ml = 0.25 L.
So, number of moles of nitrogen gas initially = 720×0.25R\dfrac{720\times 0.25}{R}
Now, since the two gases are mixed in a closed container. So,
Initial moles of nitrogen + initial moles of oxygen = final number of moles of the mixture
Therefore, we can write that, number of moles = 700×0.2R+720×0.25R=(700×0.2)+(720×0.25)R\dfrac{700\times 0.2}{R}+\dfrac{720\times 0.25}{R}=\dfrac{\left( 700\times 0.2 \right)+\left( 720\times 0.25 \right)}{R}
Now, the volume of the flask is 1 litre that means the final volume of the mixture is also 1 L.
And we have already calculated the number of moles and temperature is constant.
So, final pressure of the mixture will be = P = nRV\dfrac{nR}{V}
Putting the values in the above relation, we will get-
P=P = (700×0.2)+(720×0.25)R×R1=3201=320\dfrac{\dfrac{\left( 700\times 0.2 \right)+\left( 720\times 0.25 \right)}{R}\times R}{1}=\dfrac{320}{1}=320
So, the final pressure of the mixture = 320 mm

Therefore, the correct answer is option [B] 320.

Note: To solve this, we can also use Boyle’s law. The law describes how pressure and volume of a gas are indirectly proportional to each other at constant temperature i.e. increasing the volume decreases the pressure and vice versa. Mathematically, we can write that-
PV=constantPV=constant
Here, P and V denote pressure and volume respectively. Using this we can calculate the partial pressure and then using Dalton's law of partial pressure, we can find out the required answer.